A straight line in the complex plane

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Homework Help Overview

The discussion revolves around the properties of complex numbers and their representation in the complex plane, specifically focusing on the equation w = sz + tz* + r = 0 and its implications for straight lines in that context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether the equation w - w* = (s - t*)z + (t - s*)z* + r - r* = 0 can also represent a complex straight line, similar to the original equation. There is a focus on understanding the conditions under which these equations hold true.

Discussion Status

Some participants are seeking clarification on the nature of the equations and their forms, while others are suggesting methods to express the variables in real terms to analyze the equations further. There is an ongoing exploration of the definitions and properties of complex straight lines.

Contextual Notes

Participants note that r, s, and t are non-zero complex numbers, and z is expressed as a complex variable z = x + iy, where x and y are real numbers.

rajeshmarndi
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Homework Statement
w+w* = (s+t*)z + (t+s*)z* + r+r* = 0, is a straight line. Then couldn't find how does
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0, is also a straight line.
Relevant Equations
w = sz+tz*+r=0
sz+tz*+r=0=say w

so w* = s*z* + t*z + r*=0

Now ,
w+w* = (s+t*)z + (t+s*)z* + r+r* = 0
= p*z + pz* + k = 0...eq(1) ( k is a constant or twice real part of w)
which is in complex straight line equation form i.e ab* + a*b + c = 0 ( a,b are complex number and c a real number.

Now, again,
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0

I couldn't understand, in the solution, how this is also termed as a complex straight line like eq(1).
Since when this is worked out, it comes to be as,

q*z - qz* + id = 0 ( since r-r* will give imaginary number)

This is not in the form of a complex straight line equation.

Thanks.
 
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rajeshmarndi said:
sz+tz*+r=0=say w

so w* = s*z* + t*z + r*=0

Now ,
w+w* = (s+t*)z + (t+s*)z* + r+r* = 0
= p*z + pz* + k = 0...eq(1) ( k is a constant or twice real part of w)
which is in complex straight line equation form i.e ab* + a*b + c = 0 ( a,b are complex number and c a real number.

Now, again,
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0

I couldn't understand, in the solution, how this is also termed as a complex straight line like eq(1).
Since when this is worked out, it comes to be as,

q*z - qz* + id = 0 ( since r-r* will give imaginary number)

This is not in the form of a complex straight line equation.

Thanks.

What, in plain language, is the statement of the problem? Is it "prove that w+w*=0 describes a straight line in the complex plane"? Is it something else? I cannot figure out what you want.
 
Ray Vickson said:
What, in plain language, is the statement of the problem?
I want to know,
given w= sz+tz*+r=0

Is
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0
also a complex straight line?

[edit: r,s,t are non-zero complex number and z=x+iy (x,y ε R) ]
 
Last edited:
rajeshmarndi said:
I want to know,
given w= sz+tz*+r=0

Is
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0
also a complex straight line?

[edit: r,s,t are non-zero complex number and z=x+iy (x,y ε R) ]

Write ##w## in real terms, and expand it out to see what you get. That is, write ##s=s_1+i s_2, t = t_1+i t_2, r = r_1+i r_2## and ##z = x + i y##.
 
Ray Vickson said:
Write ##w## in real terms, and expand it out to see what you get. That is, write ##s=s_1+i s_2, t = t_1+i t_2, r = r_1+i r_2## and ##z = x + i y##.
##w## becomes,

##w = [(s_1+t_1)x + (t_2-s_2)y] + i[(s_2+t_2)x + (s_1-t_1)y+r_2]=0##

So the real terms of ##w## is ## [(s_1+t_1)x + (t_2-s_2)y]=0##
 

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